In my first answer I have described how you can find the exact solution for the 2 zeros and the 2 pole frequencies (which are identical to the wanted break frequencies).
However, here is a good approach which should be sufficient for the shown circuit.
In principle, I follow the way as outlined already by Dave Tweed´s answer: Simplification of the circuit.
In the present case, you can create three different (simplified) circuits of first order only which easily can be analysed.
1.) For the first rising region of the transfer function the high pass part with C1 is responsible (C2 causes the falling part and can be neglected). Furthermore, for very low frequencies (including DC) the gain Ao=1+R3/R2 is assumed to be not much larger than unity which is the possible minimum.
Hence, for acceptable filtering it is assumed that R2>>R3.
As an equivalent diagram for the lower frequency range (without C2 and R2) we arrive at a circuit with only the three components R1, R3 and C1. It is a simple task to find the relevant time constants (invers to the corresponding break frequencies):
Using your indices, we thus find T2=(R1+R3)C1 and T1=R1C1.
2.) Above the frequency f1 the capacitor C1 is not effective any more (and the capacitor C2 is assumed to be not yet effective). Hence, we have a simple non-inverting amplifier with the gain (maximum of the transfer function) Amax= 1+R3/Rp with Rp=R1||R2.
3.) For rising frequencies, the low pass part with capacitor C2 becomes effective (C1 is considered as a short). Hence, the feedback path consists of R3||C2 and Rp only.
The time constant T3 (pole frequency) can be derived as T3=R3C2 and the last break frequency (zero) is determined by T4=R3C2/(1+R3/Rp).
Finally, it is to be noted, that all results are in agreement with the values given in the scetched BODE diagram. This can be verified using the well-known relations for a 20dB gain slope (as used in the graph with G1/G2=f1/f2).
Final remark: Thus, it can be concluded that the information contained in the scetched BODE diagram (break frequencies) also are only approximations.
the op-amp is correctly wired up with in an inverting circuit configuration. because of the negative feedback through passive components, the "-" terminal is a virtual ground. the node equations (\$V_2\$ is the voltage at the node where are \$R_1\$, \$R_2\$, \$C_4\$, and \$C_3\$ are connected) are:
$$ \left(\frac{1}{R_1} + \frac{1}{R_2} + sC_4 + sC_3 \right)V_2 - sC_4 V_\text{out} = \frac{1}{R_1} V_\text{in}$$
$$ sC_3 V_2 + \frac{1}{R_5} V_\text{out} = 0 $$
from that, i get
$$ \begin{align}
A(s) \triangleq \frac{V_\text{out}}{V_\text{in}} & = \frac{-\frac{1}{R_1} s C_3}{\frac{1}{R_5}\left(\frac{1}{R_1} + \frac{1}{R_2} + sC_4 + sC_3 \right) + (sC_4)(sC_3) } \\
\\
& = \frac{-\frac{1}{R_1 C_4} s}{\frac{1}{R_5 C_3 C_4} \left( \frac{1}{R_1} + \frac{1}{R_2} \right) + \frac{C_4 + C_3}{R_5 C_3 C_4} s + s^2 } \\
\\
& = \frac{-H \omega_0 s}{s^2 + \frac{\omega_0}{Q} s + \omega_0^2} \\
\end{align} $$
equating the corresponding coefficients...
$$ \omega_0^2 = \frac{1}{R_5 C_3 C_4} \left( \frac{1}{R_1} + \frac{1}{R_2} \right) $$
$$ \frac{\omega_0}{Q} = \frac{C_4 + C_3}{R_5 C_3 C_4} $$
$$ H \omega_0 = \frac{1}{R_1 C_4} $$
i think the intent, in the lecture notes posted in the question is that \$ \omega_0 \triangleq 2 \pi f_\text{m} \$
so let \$ C_3 = C_4 \triangleq C \$ and let \$ k \triangleq \omega_0 C \$.
then $$ \frac{1}{R_1} = H \omega_0 C $$ $$ \frac{1}{R_2} = (2Q - H) \omega_0 C $$ $$ \frac{1}{R_5} = \frac{1}{2Q} \omega_0 C $$.
so plug this in for \$R_1\$, \$R_2\$, and \$R_5\$ and see if equality in the three "corresponding coefficients" equations above is met. if so, the transfer function, as given in the question, is correct.
Best Answer
This type of passive circuit can be easily solved and expressed in a so-called low-entropy format using the fast analytical circuits techniques or FACTs. The principle is to apply the generalized transfer function formula for a second-order system. It is defined as:
\$H(s)=\frac{H_0+s(H_1\tau_1+H_2\tau_2)+s^2H_1H_{12}\tau_1\tau_{12}}{1+s(\tau_1+\tau_2)+s^2\tau_1\tau_{12}}\$
The \$\tau\$ are the natural time constants of the circuits determined when the excitation (the stimulus, \$V_1\$, is reduced to \$0\;V\$). Here, short the input source, implying that \$C_1\$ left terminal is grounded. Now, "look" at the resistance offered by the terminals of \$C_1\$ and \$C_2\$ in this condition: \$\tau_1=C_1(R_1+R_2)\$ and \$\tau_2=C_2(R_2+R_3)\$. Then, do the same but shorting \$C_1\$ and "looking" at the resistance offered by \$C_2\$. You should find \$\tau_{12}=C_2(R_3+R_2||R_1)\$. We have \$D(s)\$ now:
\$D(s)=1+s(C_1(R_1+R_2)+C_2(R_2+R_3))+s^2C_2(R_3+R_2||R_1)C_1(R_1+R_2)\$
The dc gain (\$s=0\$) is obtained by opening all caps and you have
\$H_0=0\$
The high-frequency gains \$H\$ are found by setting the corresponding energy-storing elements in their high-frequency states. For \$H_1\$ and \$H_2\$, respectively replace \$C_1\$ and \$C_2\$ by short circuits and find: \$H_1=\frac{R_2}{R_1+R_2}\$ while \$H_2=0\$. As \$H_{12}\$ implies that both caps are shorted, \$H_{12}=0\$. We have:
\$N(s)=sH_1\tau_1=s\frac{R_2}{R_2+R_1}C_1(R_1+R_2)=sR_2C_1\$
What you do are simple sketches from which you infer the above values. No equations!
The complete transfer function involving the zero at the origin is then:
\$H(s)=\frac{sR_2C_1}{1+s(C_1(R_1+R_2)+C_2(R_2+R_3))+s^2C_2(R_3+R_2||R_1)C_1(R_1+R_2)}=\frac{\frac{s}{\omega_z}}{1+\frac{s}{\omega_0Q}+(\frac{s}{\omega_0})^2}\$
If I now factor the term \$\frac{s}{\omega_z}\$ in the numerator and \$\frac{s}{\omega_0Q}\$ in the denominator then rearrange, you obtain a true low-entropy transfer function defined as:
\$H(s)=H_{00}\frac{1}{1+Q(\frac{s}{\omega_0}+\frac{\omega_0}{s})}\$
in which (this is a raw result and you can rearrange and simplify):
\$Q=\frac{\sqrt{C_2(R_3+R_2||R_1)C_1(R_1+R_2)}}{C_1(R_1+R_2)+C_2(R_2+R_3)}\$
\$\omega_0=\frac{1}{\sqrt{C_2(R_3+R_2||R_1)C_1(R_1+R_2)}}\$
\$H_{00}=\frac{R_2C_1}{C_1(R_1+R_2)+C_2(R_2+R_3)}\$
\$H_{00}\$ is the gain in the flat region.
I have captured these equations in a Mathcad sheet to show how the reference equation (read raw expression) compares with the low-entropy format.
They perfectly match. The difference is that you now have a transfer function letting you calculate the values for all components depending on how you want to tune this filter and what attenuation you want at the peak. What truly matters is the low-entropy well-ordered form which tells you what terms contribute gains (attenuation), poles and zeros. Without this arrangement, there is no way you can design your circuit to meet a certain goal. To my opinion, the FACTs are unbeatable to obtain these results in one clean shot (you would need to rework the raw reference function to obtain the form I gave). If you are designing circuits (passive or active) and need to determine transfer functions, I encourage you to acquire that skill because once you have it, you won't go back to the classical approach. If you start slowly step by step, it is quite simple actually. Complicate expressions when you master 1st-order circuits.
You can discover FACTs further here
http://cbasso.pagesperso-orange.fr/Downloads/PPTs/Chris%20Basso%20APEC%20seminar%202016.pdf
and also through examples published in the introductory book
http://cbasso.pagesperso-orange.fr/Downloads/Book/List%20of%20FACTs%20examples.pdf